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# Prove that pivot columns of matrix A form a basis for C(A).

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Prove that pivot columns of matrix A form a basis for C(A).

From "The Four Fundamental Subspaces".

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It is proven that pivot columns of matrix A form a basis for C(A).

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*Prove that pivot columns of matrix A form a basis for C(A).

(The material is from THE FOUR FUNDAMENTAL SUBSPACES. Please show each step of your solution. Thank you.)

Proof. First of all, we need the following facts

Fact 1: The dimensions of C(A) and R(A) are equal, since both are equal to the rank of A;

Fact 2:

A matrix A is in Reduced Row-Echelon Form (RREF) if it has the following properties:

1) If a row does not consist entirely of zeroes, then the first nonzero entry in the row is a 1.

2) Any rows that consist entirely of zeroes are at the bottom of the matrix, below all nonzero rows.

3) For rows not consisting of all zeroes, ...

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• BSc , Wuhan Univ. China
• MA, Shandong Univ.
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• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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